Prove the identity: $\cos 2 \alpha - \cos 3 \alpha - \cos 4 \alpha + \cos 5 \alpha = - 4 \sin \left(\dfrac{\alpha}{2}\right) \sin \alpha \cos \left(\dfrac{7 \alpha}{2}\right)$
$\begin{aligned}
LHS & = \cos 2 \alpha - \cos 3 \alpha - \cos 4 \alpha + \cos 5 \alpha \\\\
& = \left(\cos 5 \alpha + \cos 2 \alpha\right) - \left(\cos 4 \alpha + \cos 3 \alpha\right) \\\\
& = 2 \cos \left(\dfrac{5 \alpha + 2 \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - 2 \alpha}{2}\right) - 2 \cos \left(\dfrac{4 \alpha + 3 \alpha}{2}\right) \cos \left(\dfrac{4 \alpha - 3 \alpha}{2}\right) \\\\
& = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \cos \left(\dfrac{3 \alpha}{2}\right) - 2 \cos \left(\dfrac{7 \alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) \\\\
& = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \left[\cos \left(\dfrac{3 \alpha}{2}\right) - \cos \left(\dfrac{\alpha}{2}\right)\right] \\\\
& = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \left[- 2 \sin \left(\dfrac{\dfrac{3 \alpha}{2} - \dfrac{\alpha}{2}}{2}\right) \sin \left(\dfrac{\dfrac{3 \alpha}{2} + \dfrac{\alpha}{2}}{2}\right) \right] \\\\
& = - 4 \cos \left(\dfrac{7 \alpha}{2}\right) \sin \left(\dfrac{\alpha}{2}\right) \sin \left(\dfrac{2 \alpha}{2}\right) \\\\
& = - 4 \sin \left(\dfrac{\alpha}{2}\right) \sin \alpha \cos \left(\dfrac{7 \alpha}{2}\right) = RHS
\end{aligned}$
Hence proved.